(Why) Does Algebraic Topology only deal with topological spaces up to homotopy?

The motivation I was given for algebraic topology is to assign some algebraic objects as invariants to topological spaces. This way we can show that two spaces are not homeomorphic if they are assigned a different invariant. However it seems to me like in algebraic topology the invariants are always up to homotopy equivalence or maybe even only weak homotopy equivalence. This seems strange to me. Is there a reason why other kinds of algebraic invariants that can distinguish homotopy equivalent but not homeomorphic spaces not more widely studied? What is it about weak homotopy equivalence that is so special that seems to be the main focus of all algebraic topology?

• @Mark Sure but there are a lot of homotopy equivalent spaces which are not homeomorphic. So studying only invariants that distinguish only homotopy equivalent spaces won't be enough to distinguish those. May 5, 2023 at 11:37
• @Mark But arguably classification up to homotopy is less informative because the equivalence classes are “larger” and I can be presented with homotopic spaces that are nevertheless different. The optimal information is when we have classified up to both homeomorphism and homotopy May 5, 2023 at 11:38
• It's not as weak as it seems, because we study the related spaces as well. For example, we can't tell anything about the dimension of $\mathbb R^n$ just from their trivial homology/co-homology. But if we look at the punctured $\mathbb R^n,$ removing a point, distinguishes these spaces from each other. May 5, 2023 at 11:47
• I don't think these are any less studied, just that they're harder. Even when you study, say, manifolds up to homeomorphism, you have to understand them up to homotopy first. Geometric constructions like vector bundles turn out to be a fundamental homotopy-invariant, results like the h-cobordism theorem upgrade certain homotopy equivalences to homeomorphisms and are integral to classification theory and there's many invariants that take values in homotopy or (co-)homology groups without themselves being homotopy-invariant (e.g. the Kirby-Siebenmann invariant or Pontryagin classes). May 5, 2023 at 11:52
• You might want to take a look at Scorpan's "Wild World of $4$-Manifolds." Two of the major questions there are whether $4$-manifolds necessarily have a smooth structure and whether that structure is necessarily unique. (Spoiler: The answer to both is no.) The main method for resolving it is constructing algebraic invariants, such as the signature of the intersection form, that are stronger (but more specialized) than the ones covered in books like Hatcher or Milnor and Stasheff. May 5, 2023 at 12:01

Algebraic topology would like to classify spaces up to homeomorphism (as well as some other categories of equivalence, like piecewise-linear homeomorphism) rather than just up to homotopy equivalence, but the former turns out to be much harder. The reason for the focus on the latter is just that homology and cohomology are powerful invariants that are generally reasonable to compute, and their invariance under homotopy equivalence is often good enough to prove that two spaces aren't homeomorphic. They pop up in group theory and in algebraic and differential geometry, in results like the Lefschetz fixed-point theorem and the Whitehead theorem, and a lot of other places.

The analogy I have in the back of my head is to equivalence of central extensions of groups: We actually want to classify them up to isomorphism, but that turns out to be much harder than classifying them up to equivalence via group cohomology, and the latter is often good enough for particular results. Or just consider group theory itself: We'd like to classify groups completely up to isomorphism, but that's hard to do, and there are a lot of other interesting questions along the way.

But as useful as homology and cohomology are, they're not the full extent of algebraic topology. The classification of $$3$$-dimensional lens spaces $$L(p, q)$$ up to homotopy equivalence is straightforward enough, but determining them up to homeomorphism is more complicated. (In particular, they give examples of spaces that are homotopy equivalent but not homeomorphic, beyond trivial ones like contractible spaces.) Historically, this was done with more complicated algebraic invariants like Reidemeister torsion, and that led to a variety of results in obstruction theory, higher cohomology operations, and surgery theory. I could spend pages talking about more algebraic topology constructions beyond (simplicial, cellular, etc.) homology and cohomology, so I'll limit myself to mentioning the cobordism ring, Donaldson's theorem on $$4$$-manifolds, and Milnor's construction of an exotic $$S^7$$ by the Pontryagin class.

These sorts of invariants are squarely part of algebraic topology, but they're not usually covered in introductory courses like Hatcher due to the algebraic and topology prerequisites involved; besides, at that point, you're firmly in algebraic topology (or geometric topology, but at least topology) territory, and there are fewer broadly applicable topics like general cohomology.

If you are clever you can sometimes use homotopy theory to give you complete homeomorphic information. The answer of @anomaly alludes to one example, but here is an even simpler example: a proof that if $$\mathbb R^m$$ is homeomorphic to $$\mathbb R^n$$ then $$m=n$$.

Of course $$\mathbb R^m$$ and $$\mathbb R^n$$ are homotopy equivalent, in fact both are contractible, so that does not help.

Instead we make special use of relative homology, which is a homotopy invariant in the category of topological pairs.

Suppose that there did exist a homeomorphism $$f : \mathbb R^m \to \mathbb R^n$$. Choosing any $$p \in \mathbb R^m$$, and letting $$q=f(p)$$, we obtain a homeomorphism $$f : (\mathbb R^m,\mathbb R^m - \{p\}) \to (\mathbb R^n,\mathbb R^n - \{q\})$$ in the category of pairs. Therefore, $$f$$ is also a homotopy equivalence in the category of pairs. Relative homology is a homotopy functor of the category of pairs, and so we obtain an induced isomorphism of relative homology groups (i.e. an isomorphism of "local homology groups"): $$f_* : H_i(\mathbb R^m,\mathbb R^m - \{p\}) \to H_i(\mathbb R^n,\mathbb R^n - \{q\})$$ It follows that $$m=n$$, because of the following calculation: for any point $$a \in \mathbb R^j$$ we have $$H_i(\mathbb R^j,\mathbb R^j - \{a\}) = \begin{cases} \mathbb Z \quad\text{if i=j} \\ 0 \quad\text{if i \ne j} \end{cases}$$